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प्रश्न
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove the following:
`(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
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उत्तर
LHS = `(sec θ - tan θ)/(sec θ + tan θ )`
= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`
= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`
= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`
= 1 + 2 tan2θ − 2 sec θ. tan θ
= R.H.S.
Hence proved.
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